import java.util.LinkedList;

import sun.reflect.generics.tree.VoidDescriptor;

/*
 * @lc app=leetcode.cn id=72 lang=java
 *
 * [72] 编辑距离
 *
 * https://leetcode-cn.com/problems/edit-distance/description/
 *
 * algorithms
 * Hard (54.80%)
 * Likes:    378
 * Dislikes: 0
 * Total Accepted:    17.6K
 * Total Submissions: 32.1K
 * Testcase Example:  '"horse"\n"ros"'
 *
 * 给定两个单词 word1 和 word2，计算出将 word1 转换成 word2 所使用的最少操作数 。
 *
 * 你可以对一个单词进行如下三种操作：
 *
 *
 * 插入一个字符
 * 删除一个字符
 * 替换一个字符
 *
 *
 * 示例 1:
 *
 * 输入: word1 = "horse", word2 = "ros"
 * 输出: 3
 * 解释:
 * horse -> rorse (将 'h' 替换为 'r')
 * rorse -> rose (删除 'r')
 * rose -> ros (删除 'e')
 *
 *
 * 示例 2:
 *
 * 输入: word1 = "intention", word2 = "execution"
 * 输出: 5
 * 解释:
 * intention -> inention (删除 't')
 * inention -> enention (将 'i' 替换为 'e')
 * enention -> exention (将 'n' 替换为 'x')
 * exention -> exection (将 'n' 替换为 'c')
 * exection -> execution (插入 'u')
 *
 *
 */

// @lc code=start
class Solution {
    public int minDistance(String word1, String word2) {
        int len1=word1.length(), len2 = word2.length();
        int[][] dp = new int[len1+1][len2+1];
        for(int i = 0; i <= len1; i++) {
            dp[i][0] = i;
        }
        for(int i = 0; i <= len2; i++) {
            dp[0][i] = i;
        }

        for(int i =1; i <= len1; i++) {
            for(int j=1; j <= len2; j++) {
                if(word1.charAt(i-1) == word2.charAt(j-1)) {
                    int min = Math.min(dp[i-1][j], dp[i][j-1]);
                    dp[i][j] = Math.min(min+1, dp[i-1][j-1]);
                } else {
                    int min = Math.min(dp[i-1][j], dp[i][j-1]);
                    dp[i][j] = 1 + Math.min(min, dp[i-1][j-1]);
                }
            }

        }
        return dp[len1][len2];
    }

    void dfs() {
        LinkedList list = new
    }
}
// @lc code=end

